Download Mon 2011-01-24 - Mathematics

.
MA162: Finite mathematics
.
Jack Schmidt
University of Kentucky
January 24, 2011
Soon:
HW A1 is due Today, Jan 24th, 2011.
I will be in the Mathskeller from 3pm-4pm today.
HW A2 is due Monday, Jan 31st, 2011.
HW A3 is due Sunday, Feb 6th, 2011.
Exam 1 is Monday, Feb 7th, 5:00pm-7:00pm.
Today: Ch 2.1 Systems of linear equations
2.1: Systems of equations
Supervising is hard. Workers get paid, but they need work to do
You need to pick projects that keep them working
It takes 15 minutes to assemble a MintyBoost,
and 5 minutes of work to pack it
It takes 10 minutes to assemble a TV-B-Gone,
but 5 minutes of work to pack it
Your assembly crew can work 40 hours per week,
your packaging crew can work 15 hours per week
So, how do you keep them busy?
2.1: Solving one with x and y
We’ll try one without words first:
{
4x + 3y = 10
7x − 2y = 3
Two variables, but two demands
x = 4 and y = −2 works just fine in the first one:
4(4) + 3(−2) = 16 − 6 = 10
But we aren’t satisfied,
7(4) − 2(−2) = 28 + 4 = 32 ̸= 3
How can we make both equations true at the same time?
2.1: How to solve it? Try balance
One way is to solve for one of the variables, and substitute back in
In this case, none of the variables seems easy
We can also “balance” the equations:
multiply the top by 2 and the bottom by 3
{
{
8x + 6y = 20
4x + 3y = 10
2·top
−−−−−−→
3·bot
21x − 6y = 9
7x − 2y = 3
Now if we add 20 and 9 together, we get 29. Easy.
But 20 and 9 have fancy names too, we can add them
2.1: Balanced equations become easier
Let’s add the fancy names too
{
8x + 6y = 20
(leave it alone)
−−−−−−−−−−−−−−→
Add
top
to the bottom
21x − 6y = 9
{
8x + 6y = 20
29x + 0y = 29
But now the second equation is silly easy, x = 1
{
{
8x + 6y = 20
8x + 6y = 20
(leave it alone)
−−−−−−−−−−−→
Divide by 29
x + 0y = 1
29x + 0y = 29
The first equation is easier now, since x = 1, 8x is just 8
{
{
8x + 6y = 20
0x + 6y = 12
Subtract 8·bot from top
−−−−−−−−−−−−−−−−−−→
x + 0y = 1
x + 0y = 1
2.1: Finishing up (backsolving)
Finishing up by dividing 6y = 12 by 6 to get y = 2
{
{
0x + 6y = 12
0x + y = 2
Divide top by 6
−−−−−−−−−−−−−→
x + 0y = 1
x + 0y = 1
We don’t need to write “0x” or “0y”
{
0x + y = 2
−−−−−−−−−−−−−→
x + 0y = 1
{
x
y =2
=1
Probably ought to check it worked: (x = 1, y = 2)
{
{
4x + 3y = 10
4(1) + 3(2) = 4 + 6 = 10
Plug in
−−−−−−−−→
7x − 2y = 3
7(1) − 2(2) = 7 − 4 = 3
✓
2.1: Example 1: Keeping the troops moving
15 minutes to assemble the MintyBoost, 5 minutes to pack it
10 minutes to assemble the TV-B-Gone, 5 minutes to pack it
40 hours is 40 · 60 = 2400 minutes of assembly
15 hours is 15 · 60 = 900 minutes of packing
As a system of equations:
{
15x + 10y = 2400
5x + 5y = 900
assembly
packing
We finish x MintyBoosts and y TV-B-Gones
2.1: Example 1: Just the equations
So we try to solve it: {
15x + 10y = 2400
5x + 5y = 900
{
5x + 0y = 600
Subtract 2·bot from top
−−−−−−−−−−−−−−→
5x + 5y = 900
{
5x + 0y = 600
Subtract top from bot
−−−−−−−−−−−−−−→
0x + 5y = 300
{
1x + 0y = 120
Divide by 5
−−−−−−−−−−−−−−→
0x + 1y = 60
So we ask our workers to make
x = 120 MintyBoosts and
y = 60 TV-B-Gones each week
2.1: Middle management adds more constraints
Word comes down that you need to make
twice as many MintyBoosts as TV-B-Gones
They don’t care about your need to keep the workers busy
Now we must solve:


15x + 10y = 2400
5x + 5y = 900


x − 2y =
0
Can we do it?
You bet your minty boots we can!
2.1: Impossible mission
Ok, what about this system:
{
x+ y =4
x+ y =8
People rarely are so direct about asking the impossible
{
2x + 2y = 8
x+ y =8
But with two equations can the impossible really be disguised?
{
5x + 5y = 20
4x + 4y = 32
2.1: The impossible game, k?
Ok, so maybe if we have one equation and part of another,
we can make it impossible
{
x+ y= 4
What value of k makes it impossible?
7x + ky = 56
Another:
{
3x − 7y = 13
6x + ky = 38
What value of k makes it impossible?